Leetcode 120. Triangle

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#动态规划

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本文介绍了一种使用动态规划求解三角形最小路径和问题的方法。通过从底部向上迭代计算每一步到顶部的最短路径，最终得出整个三角形的最小路径总和。此算法仅需线性空间复杂度。

/**       1
 *      1   2
 *    1   2   3
 * 1    2   3   4
 * from bottom to top, we don't need to consider the last row (b/c if row # equals to 1 there's only one number just return this value),
 * start from row 3, the minimun for row(3)(0) = min(row(4)(0), row(4)(1)), row(3)(1) = min(row(4)(1), row(4)(2)), ...
 * therefore, we just need O(n) space, initial the array by coping the values from the last row of the triangle
 */ 
public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle.size() == 0) return 0;
        if (triangle.size() == 1) return triangle.get(0).get(0);
        
        int[] mins = new int[triangle.get(triangle.size()-1).size()];
        for (int k=0; k<triangle.get(triangle.size()-1).size(); k++) 
            mins[k] = triangle.get(triangle.size()-1).get(k);
            
        int row_index = triangle.get(triangle.size()-2).size();
        for (int i=triangle.size()-2; i>=0; i--) {
            for (int j=0; j<row_index; j++) 
                mins[j] = Math.min(triangle.get(i).get(j)+mins[j], triangle.get(i).get(j)+mins[j+1]);
            row_index--;
        }
        return mins[0];
    }
}